Why LAG and LEAD for row comparison in PostgreSQL? - Purpose & Use Cases

The Big Idea

Discover how a simple SQL trick can save you hours of tedious data comparison!

The Scenario

Imagine you have a list of daily sales numbers on paper. To find out how sales changed from one day to the next, you have to look at each day, then flip back or forward to the previous or next day's number manually.

The Problem

This manual way is slow and tiring. You might lose your place, make mistakes copying numbers, or miss some days. It's hard to compare rows quickly when you have hundreds or thousands of them.

The Solution

The LAG and LEAD functions in SQL let you peek at the previous or next row's data right inside your query. This means you can easily compare values from different rows without extra work or confusion.

Before vs After

✗ Before

SELECT date, sales FROM daily_sales;
-- Then manually compare each row with the previous or next

✓ After

SELECT date, sales, LAG(sales) OVER (ORDER BY date) AS prev_day_sales, LEAD(sales) OVER (ORDER BY date) AS next_day_sales FROM daily_sales;

What It Enables

It enables quick and accurate comparisons between rows, unlocking insights like trends and changes over time with just one simple query.

Real Life Example

A store manager can instantly see if sales increased or dropped compared to the previous day, helping make better decisions fast.

Key Takeaways

Manually comparing rows is slow and error-prone.

LAG and LEAD let you access previous or next rows easily.

This makes analyzing trends and changes simple and reliable.

Practice

1. What is the main purpose of the LAG function in PostgreSQL?

easy

A. To sort rows in ascending order

B. To access data from the next row in the same result set

C. To delete duplicate rows

D. To access data from a previous row in the same result set

Solution

1. Step 1: Understand LAG function behavior

   The LAG function returns the value from a previous row based on the current row's position.

2. Step 2: Compare with other options

   LEAD accesses the next row, sorting and deleting duplicates are unrelated to LAG.

3. Final Answer:

   To access data from a previous row in the same result set -> Option D

4. Quick Check:

   LAG = previous row data access [OK]

Hint: LAG looks backward, LEAD looks forward in rows [OK]

Common Mistakes:

• Confusing LAG with LEAD
• Thinking LAG sorts rows
• Assuming LAG deletes duplicates

2. Which of the following is the correct syntax to use LEAD to get the next row's salary ordered by employee_id?

easy

A. LEAD(salary) OVER (ORDER BY employee_id)

B. LEAD(salary ORDER BY employee_id)

C. LEAD(salary) ORDER BY employee_id

D. LEAD(ORDER BY salary) OVER (employee_id)

Solution

1. Step 1: Recall correct LEAD syntax

   The correct syntax uses LEAD(column) OVER (ORDER BY column) to specify the column and ordering.

2. Step 2: Check each option

   LEAD(salary) OVER (ORDER BY employee_id) matches the correct syntax; others misuse parentheses or clause order.

3. Final Answer:

   LEAD(salary) OVER (ORDER BY employee_id) -> Option A

4. Quick Check:

   LEAD(column) OVER (ORDER BY column) [OK]

Hint: LEAD/LAG need OVER with ORDER BY inside parentheses [OK]

Common Mistakes:

• Placing ORDER BY outside OVER clause
• Missing parentheses around ORDER BY
• Putting ORDER BY inside function parameters

3. Given the table sales with columns month and revenue, what will this query return?

SELECT month, revenue, LAG(revenue) OVER (ORDER BY month) AS prev_revenue FROM sales;

medium

A. Each row shows only current month revenue

B. Each row shows current month revenue and next month's revenue

C. Each row shows current month revenue and previous month's revenue

D. Query will cause a syntax error

Solution

1. Step 1: Understand LAG with ORDER BY month

   The LAG(revenue) returns the revenue from the previous month ordered by month.

2. Step 2: Interpret query output

   Each row shows current month revenue and previous month's revenue in prev_revenue. The first row's prev_revenue will be NULL.

3. Final Answer:

   Each row shows current month revenue and previous month's revenue -> Option C

4. Quick Check:

   LAG = previous row value [OK]

Hint: LAG returns previous row's value based on ORDER BY [OK]

Common Mistakes:

• Confusing LAG with LEAD
• Expecting next month's revenue instead
• Thinking query causes error

4. Identify the error in this query:

SELECT employee_id, salary, LEAD(salary) ORDER BY employee_id FROM employees;

medium

A. ORDER BY should be inside LEAD() function

B. Missing OVER() clause for LEAD function

C. LEAD cannot be used with salary column

D. employee_id must be in GROUP BY clause

Solution

1. Step 1: Check LEAD function syntax

   LEAD requires an OVER() clause to specify ordering and partitioning.

2. Step 2: Analyze query structure

   The query uses LEAD(salary) without OVER(), causing a syntax error.

3. Final Answer:

   Missing OVER() clause for LEAD function -> Option B

4. Quick Check:

   Window functions need OVER() [OK]

Hint: Always use OVER() with LEAD/LAG [OK]

Common Mistakes:

• Omitting OVER() clause
• Placing ORDER BY outside OVER()
• Confusing window functions with aggregates

5. You want to find all employees whose salary increased compared to the previous employee ordered by hire_date. Which query correctly uses LAG to achieve this?

hard

A. SELECT employee_id, salary FROM (SELECT employee_id, salary, LAG(salary) OVER (ORDER BY hire_date) AS prev_salary FROM employees) t WHERE salary > prev_salary;

B. SELECT employee_id, salary FROM employees WHERE salary > (SELECT LAG(salary) FROM employees ORDER BY hire_date);

C. SELECT employee_id, salary, LAG(salary) OVER (ORDER BY hire_date) AS prev_salary FROM employees WHERE salary > prev_salary;

D. SELECT employee_id, salary, LAG(salary) OVER (ORDER BY hire_date) AS prev_salary FROM employees HAVING salary > prev_salary;

Solution

1. Step 1: Understand LAG usage for filtering

   Window functions like LAG cannot be used directly in WHERE clause; use subquery or CTE to compute then filter.

2. Step 2: Analyze options

   SELECT employee_id, salary FROM (SELECT employee_id, salary, LAG(salary) OVER (ORDER BY hire_date) AS prev_salary FROM employees) t WHERE salary > prev_salary; correctly uses subquery. The second option's subquery LAG lacks OVER clause (syntax error). The third uses alias in WHERE (invalid). The fourth uses HAVING without GROUP BY (invalid).

3. Final Answer:

   SELECT employee_id, salary FROM (SELECT employee_id, salary, LAG(salary) OVER (ORDER BY hire_date) AS prev_salary FROM employees) t WHERE salary > prev_salary; -> Option A

4. Quick Check:

   Subquery to filter on window functions [OK]

Hint: Use subquery to filter on window function results [OK]

Common Mistakes:

• Using alias in WHERE clause
• Misusing HAVING without GROUP BY
• Incorrect subquery for LAG